The hasFeature method is not used for this sort of thing. Try this:
if(typeof gadgets.util.escapeString == 'function') {
.....
}
On Jun 26, 2008, at 6:19 AM, Daniel wrote:
>
> Hi!
>
> Could you get the correct output with that code? I've tried it on hi5
> and it gave me "false" but Hi5 supports it.
>
> Best regards,
> Daniel Botelho
>
> On Jun 25, 8:34 pm, allano <[EMAIL PROTECTED]> wrote:
>> hi!
>>
>> Try to do it without "()" at the end of feature name. Try something
>> like this :
>>
>> gadgets.util.hasFeature("gadgets.util.escapeString");
>>
>> regards
>>
>> On Jun 25, 7:42 am, Daniel <[EMAIL PROTECTED]> wrote:
>>
>>> Hi!
>>
>>> How can I know if container supports gadgets.util.escapeString() ?
>>> I'm
>>> trying to use:
>>> "gadgets.util.hasFeature(gadgets.util.escapeString())" but this
>>> always
>>> return false.
>>
>>> Best regards,
>>> Daniel Botelho
>>
>>
> >
Paul Lindner
[EMAIL PROTECTED]
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups
"OpenSocial Application Development" group.
To post to this group, send email to [email protected]
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at
http://groups.google.com/group/opensocial-api?hl=en
-~----------~----~----~----~------~----~------~--~---
